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Question

A rod of mass M and length L is hinged at its one end and carries a particle of mass m at its lower end. A spring of force constant k1 is installed at distance a from the hinge and another of force constant, k2 at a distance b as shown in the figure. The whole arrangement rests on a smooth horizontal table top. Find the frequency of vibration.
773570_31db064b432d4e14b5c643ad0cda2b2f.png

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Solution

A is the point where the rod is hinged. Let θ be the angular displacement from mean position. The extension of spring connected at point B is

=bθ (θ is so small that we take it as linear displacement)

So, restoring force at B =k2bθ

Moment of this force about point A =k2bθ.b=k2b2θ(b is the distance of force from A)

The contraction of spring connected at point c is =aθ

So, restoring force is=k1aθ

And, the moment of force about A is =k1aθ.a=k1a2θ (a is the distance of point A of restoring force)

So, total restoring force=(k1a2+k1b2)θ

Now the total moment of inertia of the system about point A is

I= Moment of inertiaabout point A + Moment of inertial of mass m about A

=ml23+ml2

So, the angular acceleration, d2θdt2=(k1a2+k1b2)θI=(k1a2+k1b2)θl2(m3+m)

d2θdt2+(k1a2+k1b2)θl2(m3+m)=0

So, agular frequency, ω=  (k1a2+k1b2)l2(m3+m)

and frequency, f=ω2π=12π  (k1a2+k1b2)l2(m3+m)


948431_773570_ans_5c367a5b8faa4697b46de4f5e3e8a9a5.png

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